anonymous
  • anonymous
Simplify the rational expression. State any restrictions on the variable.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\frac{ n^4-10n^2+24 }{ n^4-9^2+18 }\]
math&ing001
  • math&ing001
First lets find the restrictions. For that we solve for n : n^4 - 9^2 + 18 = 0
anonymous
  • anonymous
Than what? @math&ing001

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anonymous
  • anonymous
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anonymous
  • anonymous
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anonymous
  • anonymous
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anonymous
  • anonymous
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anonymous
  • anonymous
@Hero
anonymous
  • anonymous
@satellite73
Hero
  • Hero
Let \(n^2 = y\) then try simplifying the expression.
anonymous
  • anonymous
Please help me I know its \[\frac{ n^2-4 }{ n^2-3}\] but i dont know the other half for my answer, so it's between A-C for the answer
anonymous
  • anonymous
(n+2)(n−2)(n2−6) (n2−3)(n2−6)
Hero
  • Hero
How do you know it's \(\dfrac{n^2 - 4}{n^2 - 3}\)?
anonymous
  • anonymous
(n+2)(n−2)(n2−6) (n2−3)(n2−6)
Hero
  • Hero
Multiply it back out to see if what you get matches the original expression.
Hero
  • Hero
Also the original expression you posted likely has a typo in it somewhere.
anonymous
  • anonymous
How do i do that?
anonymous
  • anonymous
@timo86m
anonymous
  • anonymous
@Astrophysics
anonymous
  • anonymous
@Michele_Laino can you please help me finish my math problem?
anonymous
  • anonymous
@mathmate
anonymous
  • anonymous
I just need help with the end of the problem. I think the answer is B though
mathmate
  • mathmate
@EllenJaz17 Please confirm that the problem is indeed: \(\Large \frac{ n^4-10n^2+24 }{ n^4-9\color{red}{n}^2+18 } \)
anonymous
  • anonymous
yes
anonymous
  • anonymous
Yes it is @mathmate
Michele_Laino
  • Michele_Laino
hint: we can factorize the denominator, using the same procedure for numerator, so we can write this: \[{n^4} - 9{n^2} + 18 = \left( {{n^2} - 6} \right)\left( {{n^2} - 3} \right)\]
anonymous
  • anonymous
(n+2)(n−2)(n2−6) (n2−3)(n2−6)
anonymous
  • anonymous
\[\frac{ n^2-4 }{ n^2-3 }\]
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
I know thats the answer to the first half but i need to find out the second half
anonymous
  • anonymous
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anonymous
  • anonymous
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anonymous
  • anonymous
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anonymous
  • anonymous
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Michele_Laino
  • Michele_Laino
the denominator is equal to zero when these two conditions hold: \[\begin{gathered} {n^2} - 6 = 0 \hfill \\ \hfill \\ {n^2} - 3 = 0 \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
please remember, that we can not divide by zero
anonymous
  • anonymous
correct
Michele_Laino
  • Michele_Laino
please solve those quadratic equations for n, what do you get?
anonymous
  • anonymous
6 and 3?
Michele_Laino
  • Michele_Laino
are you sure? we have 6^2= 36, and 3^2=9
anonymous
  • anonymous
No im not sure im just guessing
Michele_Laino
  • Michele_Laino
hint: the solution to this equation: \[{n^2} - k = 0\] are: \[n = \sqrt k ,\quad n = - \sqrt k \]
Michele_Laino
  • Michele_Laino
solutions*
Michele_Laino
  • Michele_Laino
since: \[{\left( {\sqrt k } \right)^2} = {\left( { - \sqrt k } \right)^2} = {n^2}\]
Michele_Laino
  • Michele_Laino
being k a positive number or k=0
anonymous
  • anonymous
So is it answer B?
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Michele_Laino
  • Michele_Laino
no, I don't think so, sorry!
anonymous
  • anonymous
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Michele_Laino
  • Michele_Laino
hint: what are the solution of this equation: \[{n^2} - 3 = 0\] ?
anonymous
  • anonymous
So that only leaves A & C correct? because D isn't \[\frac{ n^2-4 }{ n^2-3 }\]
Michele_Laino
  • Michele_Laino
solutions*
anonymous
  • anonymous
\[n^{2}=3?\]
Michele_Laino
  • Michele_Laino
yes! and what is n=...?
anonymous
  • anonymous
\[n \neq \pm \sqrt{6}\pm \sqrt{3}?\]
Michele_Laino
  • Michele_Laino
from this equation: \[{n^2} - 3 = 0\] I get \[n = \pm \sqrt 3 \] am I right?
anonymous
  • anonymous
yes
Michele_Laino
  • Michele_Laino
ok! so we have to exclude those 2 values, since when: \[n = \pm \sqrt 3 \] the denominator is zero Now do the same with the equation: \[{n^2} - 6 = 0\]
anonymous
  • anonymous
\[n=\pm \sqrt{6}\]
Michele_Laino
  • Michele_Laino
perfect, we have to exclude those values too, since when: \[n = \pm \sqrt 6 \] the denominator is zero. So what is the right option?
anonymous
  • anonymous
A
Michele_Laino
  • Michele_Laino
that's right! :)
anonymous
  • anonymous
Thank you so much for teaching me how to get to the answer!
Michele_Laino
  • Michele_Laino
:)

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